Question

Let f be a polynomial function such that $f\left(3x\right)=f^{\prime }\left(x\right)\cdot f^{\prime \prime }\left(x\right)$, for all $x\in R$. Then.

• A. $f^{\prime \prime }\left(2\right)-f^{\prime }\left(2\right)=0$
• B. $f\left(2\right)+f^{\prime }\left(2\right)=28$
• C. $f^{\prime \prime }\left(2\right)-f\left(2\right)=4$
• D. $f\left(2\right)-f^{\prime }\left(2\right)+f^{\prime \prime }\left(2\right)=10$

Answer 1

The correct option is A $f^{\prime \prime }\left(2\right)-f^{\prime }\left(2\right)=0$

Let $f\left(x\right)=a_o{x}^n+a_1{x}^{n-1}+a_2{x}^{n-2}+....+a_{n-1}x+a_n$
$f^{\prime }\left(x\right)=a_{o}n{x}^{n-1}+a_1(n-1)x^{n-2}+.....+a_{n-1}$
$f^{\prime \prime }\left(x\right)=a_{o}n(n-1)x^{n-2}+a_1(n-1)(n-2)x^{n-3}+....+a_{n-2}$
$f\left(3x\right)=3^n{a}_o{x}^n+3^{n-1}{a}_1{x}^{n-1}+3^{n-2}{a}_2{x}^{n-2}+....+3a_{n-1}x+a_n$
$f^{\prime }\left(x\right)f^{\prime \prime }\left(x\right)=[a_{o}n{x}^{n-1}+a_1(n-1)x^{x-2}+....+a_{n-1}]\left[a_{o}n\right(n-1)x^{n-2}+a_1(n-1\left)\right(n-2)x^{n-3}+....+a_{n-2}]$

Comparing highest powers of $x$,
$3^n{a}_o{x}^n=a_o^2(n-1)x^{n-1+n-2}=a_o^2{n}^2(n-1)x^{2n-3}$
$\therefore 2n-3=n$
$\Rightarrow n=3$

and $3^n{a}_o=a_o^2{n}^2(n-1)$
$3^3=a_{o}9(3-1)$
$a_o=\frac{27}{18}=\frac{3}{2}$
$\therefore f\left(x\right)=\frac{3}{2}{x}^3+a_1{x}^2+a_{2}x+a_3$
$f\left(3x\right)=\frac{81}{2}{x}^3+9a_1{x}^2+3a_{2}x+a_3$
$f^{\prime }\left(x\right)=\frac{9}{2}{x}^2+2a_{1}x+a_2$
$f^{\prime \prime }\left(x\right)=9x+2a_1$

$\frac{81}{2}{x}^3+9a_1{x}^2+3a_{2}x+a_3=(\frac{9}{2}{x}^2+2a_{1}x+a_2)(9x+2a_1))$
$\frac{81}{2}{x}^3+9a_1{x}^2+3a_{2}x+a_3=\frac{81}{2}{x}^3+[9a_1+18a_1]x^2+[4a_1^2+9a_2]x+2a_1{a}_2$
Comparing coefficients,
$a{a}_1=27a_1$
$\Rightarrow a_1=0$
$3a_2=4a_1^2+9a_2=9a_2$
$\Rightarrow a_2=0$
$a_3=2a_1{a}_2\Rightarrow a_3=0$

$\therefore f\left(x\right)=\frac{3}{2}{x}^3$
$f^{\prime }\left(x\right)=\frac{9}{2}{x}^2$
$f^{\prime \prime }\left(x\right)=9x$
$f^{\prime \prime }\left(2\right)-f^{\prime }\left(2\right)-18-18=0$.